Animated Solution for Mathematics - Complex Numbers: Let $S=\{z\in \mathbb{C}:|\frac{z-6i}{z-2i}|=1\text{ and }|\frac{z-8+2i}{z+2i}|=\frac{3}{5}\}$. Then ${\sum }_{z\in S}|z{|}^2$ is equal to

Welcome, fellow traveler of the complex plane! Today, we are not just solving an equation; we are embarking on a journey to uncover the hidden geometry behind a set of complex numbers.

Often, when students see a problem involving complex numbers, they immediately reach for algebraic manipulation. But wait! Before you start expanding, look at the structure.

Complex numbers are, at their heart, points in a two-dimensional plane. Let us decode this step by step.

The First Locus

We are given the condition:

Think about what this means physically. If you have a point $z$ in the complex plane, the expression $|z-6i|$ represents the distance between $z$ and the point $6i$. Similarly, $|z-2i|$ is the distance between $z$ and $2i$.

The equation tells us that these two distances are equal. Geometrically, the set of all points equidistant from two fixed points is the perpendicular bisector of the segment connecting them.

The points $6i$ and $2i$ lie on the imaginary axis at $(0,6)$ and $(0,2)$. Their midpoint is $(0,4)$, and the line connecting them is vertical.

Therefore, the perpendicular bisector must be the horizontal line $y=4$. This is our first major breakthrough! Any complex number $z$ in our set $S$ must have an imaginary part of $4$. We can write $z=x+4i$, where $x$ is a real number.

The Second Locus

Now, let us turn our attention to the second condition:

This looks intimidating, but we have a secret weapon: our substitution $z=x+4i$. By cross-multiplying, we get:

Substituting $z=x+4i$ into this, we get:

Grouping the real and imaginary parts, we have:

Now, we square both sides to eliminate the modulus. Recall that for any complex number $w=a+bi$, the squared modulus is $|w{|}^2=a^2+b^2$. Applying this, we get:

The Algebraic Resolution

Let us expand this carefully. $(x-8{)}^2=x^2-16x+64$. Adding $6^2=36$ gives us $x^2-16x+100$.

So, the equation becomes:

Distributing the $25$, we get:

Bringing all terms to one side, we arrive at:

Dividing by $16$ simplifies this to:

This is a beautiful quadratic! Factoring it, we find $(x-17)(x-8)=0$, which gives us two values: $x=17$ and $x=8$. Thus, our two complex numbers are $z_1=17+4i$ and $z_2=8+4i$.

The Final Calculation

We are almost at the finish line. The problem asks for the sum of the squared magnitudes, $\sum |z{|}^2=|z_1{|}^2+|z_2{|}^2$.

For $z_1=17+4i$:

For $z_2=8+4i$:

Adding these together:

We have successfully navigated the geometry and the algebra to find the answer. Remember, in JEE problems, always look for the geometric interpretation first—it often turns a complex calculation into a simple, elegant path. The final result is 385.